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Many situations in daily life refer to a series such as calling regularly aired episodes of Dora the Explorer a series. When used in mathematics, however, the term takes on a more elaborate meaning. In this lesson, the mathematical concept of a series will be defined, and the ways to find the sum of various special series will be practiced through real-world examples.

Catch-Up and Review

Here are a few recommended readings before getting started with this lesson.

Patterns
Sequences
Arithmetic Sequence

Explore

Finding the Sum of the First $100$ Natural Numbers

Consider the following applet. The flashing squares can be placed on any two numbers.

Addition of numbers from 1 to 100 in increasing and descending order

Now, take the sum of the first and last numbers of the given ones. Next, select the second from the initial number, and the second to the last number on the list. What is their sum? Then, select the third number and the third to last number. What is their sum? Continue doing so for a few more pairings.

123 + + + . . . 100 = ? 99 = ? 98 = ?

A pattern begins to emerge. How could this pattern be applied in calculating the sum of the first

100

natural numbers?

Discussion

Series

The sum of the terms of a sequence is called a series.

Example Sequence : Example Series : 2, 4, 6, 8 2 + 4 + 6 + 8

Depending on the number of terms, a series can be finite or infinite. A finite series has a finite number of terms. This means that there is a first and last term. On the other hand, an infinite series has an infinite number of terms. Therefore, there is a first term but not a last term because the terms continue to infinity.

Example Finite Series 2 + 4 + 6 + 8 Example Infinite Series 2 + 4 + 6 + 8 + 10 + \dots

If the sum of an infinite series approaches a number as

n

tends to infinity, then the series is said to converge to that number. Otherwise, the series diverges.

Concept

Sigma Notation

Sigma notation, also known as summation notation, is a compact way of expressing addition. This notation consists of four parts.

The Greek letter sigma $Σ .$ This letter indicates that the terms are added together.
The $general$ $form$ of all the terms being added, in terms of a variable. The variables $n$ and $i$ are commonly used.
The $summation$ $index$ and the $starting$ $index$ , also called the $lower$ $limit .$ They are placed below the letter sigma and are connected with an equals sign. The summation index is the variable used in the general form of the terms. The starting index is the first value that the variable takes.
The $final$ $index$ or $upper$ $limit .$ This number is placed above the letter sigma and indicates the value for the variable in the last term of the summation.

In the example below, all four parts are shown.

The meaning of each part of summation notation

The variable

n

— the summation index — only takes integer values. To write this sum explicitly, the variable

n

must be replaced with the integers from the initial value through the final value.

n = 1 \sum 4 2 n n = 1 \sum 4 2 n n = 1 \sum 4 2 n = n = 1 2 (1) + n = 2 2 (2) + n = 3 2 (3) + n = 4 2 (4) = 2 + 4 + 6 + 8 = 20

Extra

Notable Aspects of Summation Notation

There are some aspects that are worth noting.

The summation does not depend on the summation index used.
$n = 1 \sum 4 2 n = k = 1 \sum 4 2 k = i = 1 \sum 4 2 i = 20$
Sometimes a summation may involve other variables. These should not be confused with the summation index.
$n = 1 \sum 3 \frac{2 n}{k}$
Here, the summation index is $n .$ Therefore, the indicated values should only be substituted for $n$ and not for $k .$
$n = 1 \sum 3 \frac{2 n}{k} n = 1 \sum 3 \frac{2 n}{k} n = 1 \sum 3 \frac{2 n}{k} = n = 1 \frac{2 ( 1 )}{k} + n = 2 \frac{2 ( 2 )}{k} + n = 3 \frac{2 ( 3 )}{k} = \frac{2}{k} + \frac{4}{k} + \frac{6}{k} = \frac{1 2}{k}$
The initial value can be any integer less than or equal to the final index. The final index only indicates the last value to be substituted for $n$ — it does not indicate the number of terms.
$n = 8 \sum 10 3 n = n = 8 3 (8) + n = 9 3 (9) + n = 10 3 (10)$

The summation notation is not only useful for working with sums involving a large number of terms, but it can also be used to represent an infinite sum. If an infinite number of terms is to be added, the symbol

\infty

is used in the final index.

n = 1 \sum \infty \frac{1}{2 ^{n}} = \frac{1}{2 ^{1}} + \frac{1}{2 ^{2}} + \frac{1}{2 ^{3}} + \dots

Example

Loaves of Bread

Wilson's is a newly opened supermarket in the city center of Birmingham. There is a tasty bakery inside the supermarket.

On the day it first opened, only $22$ loaves of bread were sold. Each day after that, the bakery sold $5$ more loaves of breads than the previous day.

a Choose the option that represents the total number of loaves of bread sold in four days.

b Choose the option that represents the total number of loaves of bread sold in an infinite number days.

Hint

a Write the general rule for the number of loaves of bread sold daily.

b What will be the final index of the summation notation for an infinite number of days?

Solution

a The loaves of bread sold in four days will be expressed using summation notation. To accomplish that, the general rule of the series needs to be written. The number of loaves of bread sold in four days can be examined in a table.

Day	Loaves of Bread
$1$	$22$
$2$	$22 + 5$
$3$	$22 + 5 + 5$
$4$	$22 + 5 + 5 + 5$

The table shows that there is a pattern between the loaves sold each day. Therefore, these numbers can be considered as the terms of a sequence. Since the difference of

5

between the terms is constant, the terms represent an arithmetic sequence. Recall the explicit rule of an arithmetic sequence to best represent the number of loaves sold on

n^{th}

day.

a_{n} = a_{1} + (n - 1) d

Now, substitute the first term

a_{1} = 22

and the common difference

d = 5

into this explicit rule.

a_{n} = a_{1} + (n - 1) d

SubstituteII

$a_{1} = 22$ , $d = 5$

a_{n} = 22 + (n - 1) 5

Simplify right-hand side

Distr

Distribute $5$

a_{n} = 22 + 5 n - 5

CommutativePropAdd

\CommutativePropAdd

a_{n} = 5 n + 22 - 5

SubTerms

Subtract terms

a_{n} = 5 n + 17

Now with the general rule of the series in hand, the summation notation can be completed. To represent the first day, set the lower limit to

1 .

The upper limit, which represents the fourth day, is

4 .

n = 1 \sum 4 5 n + 17

This is a finite series.

b In the case of finding the number best representing number of loaves sold in an infinite number of days, the summation notation is the same as in Part A, but it has a different upper limit. Here, the final index is

\infty .

n = 1 \sum \infty 5 n + 17

Note that the pattern of the series and the number of loaves of bread sold each day do not change. This is why the only difference is the upper limit. Also, the series diverges since the sum does not converge to a number as

n

tends to infinity.

Example

Stacking Lemons in the Shape of a Square Pyramid

Magdalena is super stoked to get a job at the new supermarket. Her first task is to make a lemon display. She wants to stack the lemons in the shape of a pyramid with five square layers but first she needs to determine how many lemons she will need. She draws the following model sketch as a side view of a square pyramid which is formed with lemons.

a Choose the expression that represents the number of lemons in the pyramid Magdelana wants to make.

b Calculate the total number of lemons needed to make this pyramid.

Hint

a Write the rule for the number of lemons in each layer.

b Add the number of lemons in each layer one by one.

Solution

a To write the series using summation notation, consider first each layer of the pyramid.

A rule for the number of lemons in each layer will be written. To do so, a table will be made to identify the number of lemons in each layer.

Layer	Number of Lemons
$1$	$1^{2} = 1$
$2$	$2^{2} = 4$
$3$	$3^{2} = 9$
$4$	$4^{2} = 16$
$5$	$5^{2} = 25$

The number of lemons in each layer is equal to the square of the layer's level. Therefore, if the layer's level is

n,

and the number of lemons in the corresponding layer can be expressed as

n^{2} .

Using these characteristics, the summation notation can be partially written.

\sum n^{2}

Finally, the limits of the summation notation will be determined. Refer to the table to see that the initial value for the layers is

1 .

This means that the lower limit of the notation is

1,

and because the last value for the layers is

5,

the upper limit of the notation is

5 .

n = 1 \sum 5 n^{2}

b To calculate the total number of lemons in the stack, the table from Part A can be used.

Layer	Number of Lemons
$1$	$1$
$2$	$4$
$3$	$9$
$4$	$16$
$5$	$25$

Having the number of lemons in each layer, they can simply be added.

1 + 4 + 9 + 16 + 25 = 55

A total of

55

lemons are needed to make Magdalena's pyramid display.

Pop Quiz

Calculating the Sum of the Series

Calculate the sum of the series.

Discussion

Special Series

The sum of various series can be expressed by a formula. The series that share this characteristic are called special series. The table shows some of the common ones as an explicit sum and also in a summation notation.

Series	Sum	Summation Notation
Sum of $n$ terms of $1 .$	$1 + 1 + \dots + 1$	$i = 1 \sum n 1$
Sum of first $n$ positive integers.	$1 + 2 + \dots + n$	$i = 1 \sum n i$
Sum of squares of first $n$ positive integers.	$1^{2} + 2^{2} + \dots + n^{2}$	$i = 1 \sum n i^{2}$
Sum of cubes of first $n$ positive integers.	$1^{3} + 2^{3} + \dots + n^{3}$	$i = 1 \sum n i^{3}$

The formulas for these special series can save time when determining the value of the series.

Rule

Formulas for Special Series

The table shows the formulas for some special series.

Special Series	Formula
Sum of $n$ terms of $1 .$	$i = 1 \sum n 1 = n$
Sum of first $n$ positive integers.	$i = 1 \sum n i = \frac{n ( n + 1 )}{2}$
Sum of squares of first $n$ positive integers.	$i = 1 \sum n i^{2} = \frac{n ( n + 1 ) ( 2 n + 1 )}{6}$
Sum of cubes of first $n$ positive integers.	$i = 1 \sum n i^{3} = [\frac{n ( n + 1 )}{2}]^{2}$

Proof

Sum of

n

Terms of

1

Adding the same number

n

times is the same as multiplying the number by

n .

In this case, since the number is

1,

the result of this operation is

1 \cdot n,

which equals

n .

i = 1 \sum n 1 = n times 1 + 1 + 1 + \dots + 1 = n \cdot 1 = n

Proof

Sum of First

n

Positive Integers

Mathematical induction will be used to prove the following statement.

i = 1 \sum n i = \frac{n ( n + 1 )}{2}

The first step in an inductive proof is to show that the equation holds true for

n = 1 .

i = 1 \sum n i = \frac{n ( n + 1 )}{2}

Substitute

$n = 1$

i = 1 \sum 1 i = ? \frac{1 ( 1 + 1 )}{2}

Rewrite

Rewrite $i = 1 \sum 1 i$ as $1$

1 = ? \frac{1 ( 1 + 1 )}{2}

AddTerms

Add terms

1 = ? \frac{1 ( 2 )}{2}

IdPropMult

\IdPropMult

1 = ? \frac{2}{2}

QuotOne

$\frac{a}{a} = 1$

1 = 1 ✓

The statement is true for

n = 1 .

Now, assume that the statement is true for some positive integer

k .

i = 1 \sum k i = \frac{k ( k + 1 )}{2} ✓

In the final step, the aim is to show that the statement is true for

n = k + 1 .

To do so, the above equation will be manipulated using the Properties of Equality.

i = 1 \sum k i = \frac{k ( k + 1 )}{2}

AddEqn

$LHS + (k + 1) = RHS + (k + 1)$

i = 1 \sum k i + (k + 1) = \frac{k ( k + 1 )}{2} + (k + 1)

Rewrite

Rewrite $i = 1 \sum k i + (k + 1)$ as $i = 1 \sum k + 1 i$

i = 1 \sum k + 1 i = \frac{k ( k + 1 )}{2} + (k + 1)

Simplify right-hand side

NumberToFrac

$a = \frac{2 \cdot a}{2}$

i = 1 \sum k + 1 i = \frac{k ( k + 1 )}{2} + \frac{2 ( k + 1 )}{2}

AddFrac

Add fractions

i = 1 \sum k + 1 i = \frac{k ( k + 1 ) + 2 ( k + 1 )}{2}

FactorOut

Factor out $(k + 1)$

i = 1 \sum k + 1 i = \frac{( k + 1 ) ( k + 2 )}{2}

WriteSum

Write as a sum

i = 1 \sum k + 1 i = \frac{( k + 1 ) ( k + 1 + 1 )}{2}

The left-hand side of the above equation is the sum of the first

k + 1

positive integers. The right-hand side is the expression obtained when substituting

k + 1

for

n .

Therefore, the statement holds true for all positive integers.

Proof

Sum of Squares of First

n

Positive Integers

The following statement will be proved using mathematical induction.

i = 1 \sum n i^{2} = \frac{n ( n + 1 ) ( 2 n + 1 )}{6}

For the first step, it should be shown that the equation holds true for

n = 1 .

i = 1 \sum n i^{2} = \frac{n ( n + 1 ) ( 2 n + 1 )}{6}

Substitute

$n = 1$

i = 1 \sum 1 i^{2} = ? \frac{1 ( 1 + 1 ) ( 2 ( 1 ) + 1 )}{6}

Evaluate left-hand side

Rewrite

Rewrite $i = 1 \sum 1 i^{2}$ as $1^{2}$

1^{2} = ? \frac{1 ( 1 + 1 ) ( 2 ( 1 ) + 1 )}{6}

BaseOne

$1^{a} = 1$

1 = ? \frac{1 ( 1 + 1 ) ( 2 ( 1 ) + 1 )}{6}

Evaluate right-hand side

IdPropMult

\IdPropMult

1 = ? \frac{1 ( 1 + 1 ) ( 2 + 1 )}{6}

AddTerms

Add terms

1 = ? \frac{1 ( 2 ) ( 3 )}{6}

Multiply

1 = ? \frac{6}{6}

QuotOne

$\frac{a}{a} = 1$

1 = 1 ✓

The statement is true for

n = 1 .

Now, assume that the statement is true for some positive integer

k .

i = 1 \sum k i^{2} = \frac{k ( k + 1 ) ( 2 k + 1 )}{6} ✓

Next, show that the statement is true for

n = k + 1 .

To do so, start by adding

(k + 1)^{2}

to both sides of the above equation.

i = 1 \sum k i^{2} = \frac{k ( k + 1 ) ( 2 k + 1 )}{6}

AddEqn

$LHS + (k + 1)^{2} = RHS + (k + 1)^{2}$

i = 1 \sum k i^{2} + (k + 1)^{2} = \frac{k ( k + 1 ) ( 2 k + 1 )}{6} + (k + 1)^{2}

Rewrite

Rewrite $i = 1 \sum k i^{2} + (k + 1)^{2}$ as $i = 1 \sum k + 1 i^{2}$

i = 1 \sum k + 1 i^{2} = \frac{k ( k + 1 ) ( 2 k + 1 )}{6} + (k + 1)^{2}

Add the terms on the right-hand side, then factor the numerator of the fraction.

i = 1 \sum k + 1 i^{2} = \frac{k ( k + 1 ) ( 2 k + 1 )}{6} + (k + 1)^{2}

Simplify right-hand side

NumberToFrac

$a = \frac{6 \cdot a}{6}$

i = 1 \sum k + 1 i^{2} = \frac{k ( k + 1 ) ( 2 k + 1 )}{6} + \frac{6 ( k + 1 ) ^{2}}{6}

AddFrac

Add fractions

i = 1 \sum k + 1 i^{2} = \frac{k ( k + 1 ) ( 2 k + 1 ) + 6 ( k + 1 ) ^{2}}{6}

FactorOut

Factor out $(k + 1)$

i = 1 \sum k + 1 i^{2} = \frac{( k + 1 ) [ k ( 2 k + 1 ) + 6 ( k + 1 ) ]}{6}

Distr

Distribute $k & 6$

i = 1 \sum k + 1 i^{2} = \frac{( k + 1 ) ( 2 k ^{2} + k + 6 k + 6 )}{6}

WriteSum

Write as a sum

i = 1 \sum k + 1 i^{2} = \frac{( k + 1 ) ( 2 k ^{2} + k + 3 k + 3 k + 6 )}{6}

AddTerms

Add terms

i = 1 \sum k + 1 i^{2} = \frac{( k + 1 ) ( 2 k ^{2} + 4 k + 3 k + 6 )}{6}

FactorOut

Factor out $2 k$

i = 1 \sum k + 1 i^{2} = \frac{( k + 1 ) [ 2 k ( k + 2 ) + 3 k + 6 ]}{6}

FactorOut

Factor out $3$

i = 1 \sum k + 1 i^{2} = \frac{( k + 1 ) [ 2 k ( k + 2 ) + 3 ( k + 2 ) ]}{6}

FactorOut

Factor out $(k + 2)$

i = 1 \sum k + 1 i^{2} = \frac{( k + 1 ) ( k + 2 ) ( 2 k + 3 )}{6}

WriteSum

Write as a sum

i = 1 \sum k + 1 i^{2} = \frac{( k + 1 ) ( k + 1 + 1 ) ( 2 k + 2 + 1 )}{6}

FactorOut

Factor out $2$

i = 1 \sum k + 1 i^{2} = \frac{( k + 1 ) ( k + 1 + 1 ) [ 2 ( k + 1 ) + 1 ]}{6}

The left-hand side is the sum of the squares of the first

n

positive integers. The right-hand side is the expression obtained when substituting

k + 1

for

n .

Therefore, the statement holds for all positive integers

n .

Proof

Sum of Cubes of First

n

Positive Integers

The following statement can be proved by mathematical induction.

i = 1 \sum n i^{3} = [\frac{n ( n + 1 )}{2}]^{2}

In the first step, it should be shown that the equation holds true for

n = 1 .

i = 1 \sum n i^{3} = [\frac{n ( n + 1 )}{2}]^{2}

Substitute

$n = 1$

i = 1 \sum 1 i^{3} = ? [\frac{1 ( 1 + 1 )}{2}]^{2}

Evaluate left-hand side

Rewrite

Rewrite $i = 1 \sum 1 i^{3}$ as $1^{3}$

1^{3} = ? [\frac{1 ( 1 + 1 )}{2}]^{2}

BaseOne

$1^{a} = 1$

1 = ? [\frac{1 ( 1 + 1 )}{2}]^{2}

Evaluate right-hand side

AddTerms

Add terms

1 = ? [\frac{1 ( 2 )}{2}]^{2}

IdPropMult

\IdPropMult

1 = ? (\frac{2}{2})^{2}

QuotOne

$\frac{a}{a} = 1$

1 = ? 1^{2}

BaseOne

$1^{a} = 1$

1 = 1 ✓

The statement is true for

n = 1 .

Now, assume that the statement is true for some positive integer

k .

i = 1 \sum k i^{3} = [\frac{k ( k + 1 )}{2}]^{2} ✓

Finally, to show that the statement is true for

k + 1,

the above equation will be manipulated.

i = 1 \sum k i^{3} = \frac{k ^{2} ( k + 1 ) ^{2}}{4}

AddEqn

$LHS + (k + 1)^{3} = RHS + (k + 1)^{3}$

i = 1 \sum k i^{3} + (k + 1)^{3} = \frac{k ^{2} ( k + 1 ) ^{2}}{4} + (k + 1)^{3}

Rewrite

Rewrite $i = 1 \sum k i^{3} + (k + 1)^{3}$ as $i = 1 \sum k + 1 i^{3}$

i = 1 \sum k + 1 i^{3} = \frac{k ^{2} ( k + 1 ) ^{2}}{4} + (k + 1)^{3}

Simplify right-hand side

NumberToFrac

$a = \frac{4 \cdot a}{4}$

i = 1 \sum k + 1 i^{3} = \frac{k ^{2} ( k + 1 ) ^{2}}{4} + \frac{4 ( k + 1 ) ^{3}}{4}

AddFrac

Add fractions

i = 1 \sum k + 1 i^{3} = \frac{k ^{2} ( k + 1 ) ^{2} + 4 ( k + 1 ) ^{3}}{4}

FactorOut

Factor out $(k + 1)^{2}$

i = 1 \sum k + 1 i^{3} = \frac{( k + 1 ) ^{2} ( k ^{2} + 4 ( k + 1 ) )}{4}

Distr

Distribute $4$

i = 1 \sum k + 1 i^{3} = \frac{( k + 1 ) ^{2} ( k ^{2} + 4 k + 4 )}{4}

FacPosPerfectSquare

$a^{2} + 2 a b + b^{2} = (a + b)^{2}$

i = 1 \sum k + 1 i^{3} = \frac{( k + 1 ) ^{2} ( k + 2 ) ^{2}}{4}

ProdPowII

$a^{m} b^{m} = (a b)^{m}$

i = 1 \sum k + 1 i^{3} = \frac{[ ( k + 1 ) ( k + 2 ) ] ^{2}}{4}

WritePow

Write as a power

i = 1 \sum k + 1 i^{3} = \frac{[ ( k + 1 ) ( k + 2 ) ] ^{2}}{2 ^{2}}

$\frac{a ^{m}}{b ^{m}} = (\frac{a}{b})^{m}$

i = 1 \sum k + 1 i^{3} = [\frac{( k + 1 ) ( k + 2 )}{2}]^{2}

WriteSum

Write as a sum

i = 1 \sum k + 1 i^{3} = [\frac{( k + 1 ) ( k + 1 + 1 )}{2}]^{2}

This final equation shows that the statement is true for

n = k + 1 .

Therefore, the statement holds for all positive integers

n .

Example

Number of People in a Market

Magdalena has enjoyed working at Wilson's Supermarket but feels her tasks have been too easy. She decides to challenge herself and prepare a report about the number of people shopping from $9 : 00 AM$ to $6 : 00 PM$ on a weekday. She will present this report to her coworkers.

After collecting data, she is able to make the following table which highlights the number of people and the corresponding times they are shopping.

Time Interval	Number of People Shopping
$09 : 00 - 10 : 00$	$8$
$10 : 00 - 11 : 00$	$27$
$11 : 00 - 12 : 00$	$64$
$12 : 00 - 13 : 00$	$125$
$13 : 00 - 14 : 00$	$216$
$14 : 00 - 15 : 00$	$125$
$15 : 00 - 16 : 00$	$64$
$16 : 00 - 17 : 00$	$27$
$17 : 00 - 18 : 00$	$8$

Help Magdalena calculate the total number of people shopping using the formulas for special series!

Hint

The numbers in the table can be rewritten as the cubes of some positive integers. Use this general rule while representing the series in summation notation.

Solution

To express the given numbers as a series in summation notation, a general rule needs to be found first. Begin by rewriting the integers representing the number of people shopping into exponent form and recognize any emerging pattern.

Time Intervals	Number of People Shopping
$09 : 00 - 10 : 00$	$8 = 2^{3}$
$10 : 00 - 11 : 00$	$27 = 3^{3}$
$11 : 00 - 12 : 00$	$64 = 4^{3}$
$12 : 00 - 13 : 00$	$125 = 5^{3}$
$13 : 00 - 14 : 00$	$216 = 6^{3}$
$14 : 00 - 15 : 00$	$125 = 5^{3}$
$15 : 00 - 16 : 00$	$64 = 4^{3}$
$16 : 00 - 17 : 00$	$27 = 3^{3}$
$17 : 00 - 18 : 00$	$8 = 2^{3}$

The series can be written as a sum of cubes of some positive integers. In other words, the

k^{th}

term of the series can be written as

k^{3} .

General Term : a_{k} = k^{3}

Notice that the bases of the cubes are increasing and then decreasing one by one. Since writing a common upper limit or a common lower limit is not possible here, this sum cannot be expressed with one summation notation. Therefore, the series will be written by using two separate summation notations.

2^{3} + 3^{3} + 4^{3} + 5^{3} + 6^{3} = k = 2 \sum 6 k^{3} + 5^{3} + 4^{3} + 3^{3} + 2^{3} = k = 2 \sum 5 k^{3}

To find the total sum, these two summation notations will be calculated individually and their results will be added together.

k = 2 \sum 6 k^{3} + k = 2 \sum 5 k^{3}

Now, remember the special formula for the sum of cubes of the first

n

positive integers.

k = 1 \sum n k^{3} = [\frac{n ( n + 1 )}{2}]^{2}

Keep in mind that the obtained summation notations start with

2

and not

1 .

This means that the first terms of the series

a_{1}

are missing. For the formula to be used, the initial values must be

1 .

Therefore, this value will be changed to

1,

and then the first term of each series will be subtracted.

k = 2 \sum 6 k^{3} = k = 1 \sum 6 k^{3} - a_{1} k = 2 \sum 5 k^{3} = k = 1 \sum 5 k^{3} - a_{1}

Now, the formula can be applied to the right-hand sides of the equations one at a time. Since the upper limit is

6,

begin by substituting

n = 6

into the first equation.

k = 1 \sum 6 k^{3} - a_{1}

Rewrite

Rewrite $k = 1 \sum 6 k^{3}$ as $[\frac{n ( n + 1 )}{2}]^{2}$

[\frac{n ( n + 1 )}{2}]^{2} - a_{1}

Substitute

$n = 6$

[\frac{6 ( 6 + 1 )}{2}]^{2} - a_{1}

Evaluate

AddTerms

Add terms

[\frac{6 ( 7 )}{2}]^{2} - a_{1}

Multiply

(\frac{4 2}{2})^{2} - a_{1}

CalcQuot

Calculate quotient

2 1^{2} - a_{1}

CalcPow

Calculate power

441 - a_{1}

Next, the formula can be applied one more time for the second summation notation. This time the value of

n = 5

will be substituted, since the upper limit is

5 .

k = 1 \sum 5 k^{3} - a_{1}

Rewrite

Rewrite $k = 1 \sum 5 k^{3}$ as $[\frac{n ( n + 1 )}{2}]^{2}$

[\frac{n ( n + 1 )}{2}]^{2} - a_{1}

Substitute

$n = 5$

[\frac{5 ( 5 + 1 )}{2}]^{2} - a_{1}

Evaluate

AddTerms

Add terms

[\frac{5 ( 6 )}{2}]^{2} - a_{1}

Multiply

(\frac{3 0}{2})^{2} - a_{1}

CalcQuot

Calculate quotient

1 5^{2} - a_{1}

CalcPow

Calculate power

225 - a_{1}

Before adding the obtained results, the values of the first terms need to be found. Since both of the summation notations have the same general rule

k^{3},

the value of

a_{1}

will be calculated once.

k = 1 \Rightarrow k^{3} = 1^{3} a_{1} = 1

Having the value of

a_{1} = 1,

the total sum can now be calculated.

441 - a_{1} + 225 - a_{1}

Substitute

$a_{1} = 1$

441 - 1 + 225 - 1

AddSubTerms

Add and subtract terms

664

The total number of people shopping at the supermarket between

9 : 00 AM

6 : 00 PM

on a weekday is

664 .

Magdalena can now give her report in full confidence!

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Gauss Summation

Carl Friedrich Gauss was one of the greatest mathematicians of all time. Even as a middle school student in Germany, Gauss was already trying to find the sum of the first one-hundred natural numbers.

Carl Friedrich Gauss finding the sum of the first 100 natural numbers

An observation he made was that he got

101

when adding the first number with the last number, the second number with the second to last number, and so on.

Gauss realized that each of these pairs resulted in

101 .

By using this pattern, he decided to calculate the number of pairs. Because there are

100

numbers, the number of pairs is

\frac{1 0 0}{2} = 50 .

This means that the sum of the first

100

natural numbers is equal to

50

times

101 .

50 \times 101 = 5050

By using this method, Gauss easily found the sum of the numbers from

1

100 .

This is the reason why this summation is called Gauss Summation. This method can be generalized for the sum of the first

n

natural numbers.

The number of pairs is obtained by dividing the last number

n

2 .

Also, the sum of each pair is equal to the addition of the first number

1

and last number

n,

which is

n + 1 .

Using this information, the formula can be written.

Total Sum = Number of pairs \cdot Sum of each pair

SubstituteValues

Substitute values

Total Sum = \frac{n}{2} \cdot (n + 1)

MoveRightFacToNum

$\frac{a}{c} \cdot b = \frac{a \cdot b}{c}$

Total Sum = \frac{n ( n + 1 )}{2}

The obtained formula represents the sum of first

n

positive integers.

i = 1 \sum n i = \frac{n ( n + 1 )}{2}

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